Optimal control problem for coupled time-fractional diffusion systems with final observations

ABSTRACT

In this paper, fractional optimal control problem for two-dimensional coupled diffusion system with final observation is investigated. The fractional time derivative is considered in Atangana–Baleanu sense. Constraints on controls are imposed. First, by means of the classical control theory, the existence and uniqueness of the state for these systems is proved. Then, the necessary and sufficient optimality conditions for the fractional Dirichlet problems with the quadratic performance functional are derived. Finally we give some examples to illustrate the applicability of our results. The optimization problem presented in this paper constitutes a generalization of the optimal control problem of diffusion equations with Dirichlet boundary conditions considered in recent papers to coupled systems with Atangana–Baleanu time derivatives.

KEYWORDS:

• Optimal control
• coupled time-fractional system
• Atangana–Baleanu derivative
• diffusion equations
• wave equations

2010 MATHEMATICS SUBJECT CLASSIFICATIONS:

• 46C05
• 49J20
• 93C20

1. Introduction

Let $n\in N^{\ast }$ and Ω be a bounded open subset of $R^n$ with a smooth boundary Γ of class $C^2$. For a time T>0, we set $Q=\mathrm{\Omega }\times (0,T)$ and $\mathrm{\Sigma }=\mathrm{\Gamma }\times (0,T)$.

For $y_{0,1},y_{0,2}\in H_0^1\left(\mathrm{\Omega }\right)$ and $f_1,f_2\in L^2(0,T;H^{-1}(\mathrm{\Omega }\left)\right)$, let us consider the following fractional problem for coupled diffusion system: (1.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)=f_1\left(t\right),\\ & a.e.t\in ]0,T[,\end{array}$(1.1) (1.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2\left(y\right)+y_1\left(t\right)=f_2\left(t\right),\\ & a.e.t\in ]0,T[,\end{array}$(1.2) (1.3) $y_1(x,0)=y_{0,1}\left(x\right),x\in \mathrm{\Omega },$(1.3) (1.4) $y_2(x,0)=y_{0,2}\left(x\right),x\in \mathrm{\Omega },$(1.4) (1.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma },$(1.5) where $\alpha \in (0,1)$, ${}_0^{ABC}{D}_t^{\alpha }$ is the Atangana–Baleanu fractional derivatives in the Caputo sense.

The study of fractional calculus with the non-singular kernel is gaining more and more attention. Compared with classical fractional calculus with singular kernel, non-singular kernel models can describe reality more accurately, which has been shown recently in a variety of fields such as physics, chemistry, biology, economics, signal and image processing, control, porous media, aerodynamics and so on. For example, extensive treatment and various applications of the fractional calculus with the non-singular kernel are discussed in the works of Atangana et al. [1–3], Baleanu et al. [4, 5], Caputo [6], Djida et al. [7, 8], Gomez-Aguilar et al. [9–11]. It has been demonstrated that Fractional Order Differential Equations (FODEs) with the non-singular kernel models dynamic systems and processes more accurately than FODEs with singular kernel do, and fractional controllers perform better than integer order controllers.

There are many works on fractional diffusion equations and fractional diffusion wave equations. For instance, Agrawal [12, 13] studied the solutions for a fractional diffusion-wave equation defined in a boun-ded domain when the fractional time derivative is described in the Caputo sense. Using Laplace transform and finite sine transform technique, the author obtained the general solution in terms of Mittag–Leffler functions. In [14], Mophou et al. studied by means of eigenfunctions the control problems for fractional diffusion wave equation involving Riemann–Liouville fractional derivative or order $\alpha \in (\frac{3}{2},2)$.

Integer order optimal control problems for evolution equations have been extensively studied by many authors, for comprehensive treatment of this topic we refer to the classical monograph by Lions [15] and to [16].

Extensive treatment and various applications of the fractional calculus are discussed in the works of Agrawal et al. [12, 13], Ahmad and Ntouyas [17], Bahaa et al. [18–22], Mophou [23–25], Debbouche and Nieto [26,27], Wang and Zhou [28], Tang and Ma [29], etc. It has been demonstrated that FODE models dynamic systems and processes more accurately than integer order differential equations do, and fractional controllers perform better than integer order controllers.

Optimal control of fractional diffusion equations has also been studied by several authors. For instance, in [30], Agrawal considered two problems, the simplest fractional variation problem and fractional variational problem of Lagrange. For both problems, the author developed the Euler–Lagrange type necessary conditions which must be satisfied for the given functional to be extremum. In [31], a general formulation and solution scheme for fractional optimal control problems are defined in sense of Riemann–Liouville and Caputo.

In [17], Ahmad et al. investigated the existence of solutions for fractional differential inclusions with four-point non-local Riemann–Liouville type integral boundary conditions. In [32], Bahaa presented the fractional optimal control problem for variational inequalities with control constraints. The author showed that the considered optimal control problem has a unique solution. In [33], Bahaa proposed necessary conditions for optimality in optimal control problems with dynamics by differential equations of fractional order. In [20], Bahaa proposed the fractional optimal control problem for infinite order system with control constraints. Following the same technique [21], Bahaa presented the formulation of fractional optimal control problem when the dynamic constraints of the system are given by a fractional differential system and the performance index is described with a state and a control function. In [34], Baleanu et al. gave formulation for a fractional optimal control problems when the dimensions of the state and control variables are different from each other.

In [14, 23, 24], Mophou applied the classical control theory to a fractional diffusion equation involving Riemann–Liouville fractional derivative in a bounded domain. The author showed that the considered optimal control problem has a unique solution. In [35], Mophou et al. showed that existence and uniqueness of the fractional optimal control with final observation when the dynamic constraints is described by a fractional diffusion equation involving Riemann-Liouville fractional derivative. In [25], Mophou et al. showed the initial value/boundary value problem for composite fractional relaxation equation. In [29], Tang investigated the variational formulation and optimal control of fractional diffusion equations with Caputo derivatives. We also refer to [26, 27, 36, 37] and references therein for more literature on optimal control of fractional evolution and diffusion equations.

In this paper, we consider optimal control problem for coupled diffusion system with Atangana–Baleanu fractional derivatives in the Caputo sense with final observation. The novelties of this contribution is we generalize the previous studies by Agrawal et al. [12,13] and Mophou [23, 24, 35] for fractional coupled diffusion systems which can used to describe many physical, chemical, mathematical and biological models. First, by means of the classical control theory, the existence and uniqueness of the state for these systems is proved. Fractional optimal control is characterized by the adjoint problem. By using this characterization, particular properties of fractional optimal control are proved.

This paper is organized as follows. In Section 2, we introduce some basic definitions and preliminary results. In Section 3, we give some properties of Atangana–Baleanu fractional derivatives and integration by parts. In Section 4, we formulate the fractional Dirichlet problem for diffusion equations. In Section 5, we show that our fractional optimal control problem holds and we give the optimality conditions for the optimal control. In Section 6, some illustrated examples are stated. In Section 7, we state our conclusions.

2. Preliminaries

Many definitions have been given for a fractional derivative, which include Riemann–Liouville, Grü nwald–Let-nikov, Weyl, Caputo, Marchaud and Riesz fractional derivatives. We will formulate the problem in terms of the left and right Caputo fractional derivatives which will be given later.

Definition 2.1

[2]

For a given function $x\left(t\right)\in H^1(a,b),b>a$, $\alpha \in (0,1)$, the left Atangana–Baleanu fractional derivative (AB derivative) of $x\left(t\right)$ of order α in Caputo sense ${}_a^{ABC}{D}_t^{\alpha }x\left(t\right)$ (where A denotes Atangana, B denotes Baleanu and C denotes Caputo type) with base point a is defined at a point $t\in (a,b)$ by (2.1) $\begin{array}{rl}{}_a^{ABC}{D}_t^{\alpha }x\left(t\right)& =\frac{B\left(\alpha \right)}{1-\alpha }{\int }_a^t\frac{\mathrm{d}}{\mathrm{d}s}x\left(s\right)E_{\alpha }[-\gamma (t-s{)}^{\alpha }]\mathrm{d}s,\\ & \text{(left ABCD)}\end{array}$(2.1) where $\gamma =\alpha /\left((1-\alpha )\right)$, and $B\left(\alpha \right)$ being a normalization function satisfying (2.2) $B\left(\alpha \right)=(1-\alpha )+\frac{\alpha }{\mathrm{\Gamma }\left(\alpha \right)},$(2.2) where $B\left(0\right)=B\left(1\right)=1,$ $E_{\alpha }(.)$ stands for the Mittag–Lef-fler function defined by (2.3) $E_{\alpha ,\beta }\left(z\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{z^k}{\mathrm{\Gamma }(ka+\beta )},E_{\alpha ,1}\left(z\right)=E_{\alpha }\left(z\right)z\in \mathcal{C},$(2.3) which is an entire function on the complex plane and $\mathrm{\Gamma }(.)$ denotes thee Euler's gamma function defined as $\mathrm{\Gamma }\left(z\right)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{\text{e}}^{-t}\mathrm{d}t,\mathcal{R}\left(z\right)>0.$

The Mittag–Leffler function $E_{\alpha ,\beta }\left(z\right)$ is a two-para-meter family of entire functions of z of order ${\alpha }^{-1}$. Furthermore, we recall the following Lemma.

Lemma 2.1

[2]

Let $\alpha ,\beta \in \mathbb{C}$ such that $\mathcal{R}\left(\alpha \right)>0$ and $\mathcal{R}\left(\beta \right)>0$. Then we have that (2.4) $\begin{array}{rl}\left(\frac{\mathrm{d}}{\mathrm{d}z}\right)E_{\alpha ,\beta }\left(z\right)& =\frac{1}{\alpha }\left[(1+\alpha -\beta )E_{\alpha ,\beta }\left(z\right)+E_{\alpha ,\beta -1}\left(z\right)\right],\\ & z\in \mathbb{C}.\end{array}$(2.4)

The left Atangana–Baleanu fractional derivative in Riemann–Liouville sense defined with (2.5) $\begin{array}{rl}{}_a^{ABR}{D}_t^{\alpha }x\left(t\right)& =\frac{B\left(\alpha \right)}{1-\alpha }\frac{d}{\mathrm{d}t}{\int }_a^{t}x\left(s\right)E_{\alpha }[-\gamma (t-s{)}^{\alpha }]\mathrm{d}s,\\ & \text{(left ABRD)}\end{array}$(2.5) For $\alpha =1$ in (2.5(2.5) $\begin{array}{rl}{}_a^{ABR}{D}_t^{\alpha }x\left(t\right)& =\frac{B\left(\alpha \right)}{1-\alpha }\frac{d}{\mathrm{d}t}{\int }_a^{t}x\left(s\right)E_{\alpha }[-\gamma (t-s{)}^{\alpha }]\mathrm{d}s,\\ & \text{(left ABRD)}\end{array}$(2.5) ), we consider the usual classical derivative ${\mathrm{\partial }}_t$.

The associated left AB fractional integral is also defined as (2.6) $\begin{array}{rl}{AB}_a{I}_t^{\alpha }x\left(t\right)& =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)\mathrm{\Gamma }\left(\alpha \right)}\\ & \times {\int }_a^{t}x\left(s\right)(t-s{)}^{\alpha -1}\mathrm{d}s,\text{(left ABI)}\\ & =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)}{}_a{I}_t^{\alpha }x\left(t\right),\end{array}$(2.6) where (2.7) ${}_a{I}_t^{\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_a^{t}x\left(s\right)(t-s{)}^{\alpha -1}\mathrm{d}s$(2.7) is the classical left Riemann–Liouville integral.

Notice that if $\alpha =0$ in (2.6(2.6) $\begin{array}{rl}{AB}_a{I}_t^{\alpha }x\left(t\right)& =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)\mathrm{\Gamma }\left(\alpha \right)}\\ & \times {\int }_a^{t}x\left(s\right)(t-s{)}^{\alpha -1}\mathrm{d}s,\text{(left ABI)}\\ & =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)}{}_a{I}_t^{\alpha }x\left(t\right),\end{array}$(2.6) ), we recover the initial function and if $\alpha =1$ in (2.6(2.6) $\begin{array}{rl}{AB}_a{I}_t^{\alpha }x\left(t\right)& =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)\mathrm{\Gamma }\left(\alpha \right)}\\ & \times {\int }_a^{t}x\left(s\right)(t-s{)}^{\alpha -1}\mathrm{d}s,\text{(left ABI)}\\ & =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)}{}_a{I}_t^{\alpha }x\left(t\right),\end{array}$(2.6) ) we consider the usual ordinary integral. So from the definition on [2], we recall the following definition.

Definition 2.2

[2]

For a given function $x\left(t\right)\in H^1(a,b),b>t>a$, the right Atangana–Baleanu fractional derivative of $x\left(t\right)$ of order α in Caputo sense with base point b is defined at a point $t\in (a,b)$ by (2.8) $\begin{array}{rl}{(}_t^{ABC}{D}_b^{\alpha }x\left)\right(t)& =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_t^b{E}_{\alpha }\\ & \times [-\gamma (s-t{)}^{\alpha }\frac{\mathrm{d}}{\mathrm{d}s}x\left(s\right)]\mathrm{d}s,\text{(right ABCD)}\end{array}$(2.8) and in Riemann–Liouville sense with (2.9) $\begin{array}{rl}{(}_t^{ABR}{D}_b^{\alpha }x\left)\right(t)& =-\frac{B\left(\alpha \right)}{1-\alpha }\frac{\mathrm{d}}{\mathrm{d}t}{\int }_t^{b}x\left(s\right)E_{\alpha }\\ & \times [-\gamma (s-t{)}^{\alpha }]\mathrm{d}s,\text{(right ABRD)}\end{array}$(2.9) The associated right AB fractional integral is also defined as (2.10) $\begin{array}{rl}{(}_t^{AB}{I}_b^{\alpha }x\left)\right(t)& =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)\mathrm{\Gamma }\left(\alpha \right)}\\ & \times {\int }_t^{b}x\left(s\right)(s-t{)}^{\alpha -1}\mathrm{d}s,\text{(right ABI)}\\ & =\frac{1-\alpha }{B\left(\alpha \right)}x\left(t\right)+\frac{\alpha }{B\left(\alpha \right)}\left({}_t{I}_b^{\alpha }x\right)\left(t\right),\end{array}$(2.10) where (2.11) ${}_t{I}_b^{\alpha }x\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_t^{b}x\left(s\right)(s-t{)}^{\alpha -1}\mathrm{d}s$(2.11) is the classical right Riemann–Liouville integral.

3. Some properties of AB derivatives and integration by parts

In this section, we state some important lemmas for properties of AB derivatives and integration by parts which can found in [4]. Some recent results and properties concerning this operator can be found [2] and papers therein.

Lemma 3.1

[4]

The left AB Caputo fractional derivatives and the left AB Riemann–Liouville derivative are related by the identity: (3.1) ${}_a^{ABC}{D}_t^{\alpha }x\left(t\right)={}_a^{ABR}{D}_t^{\alpha }x\left(t\right)-\frac{B\left(\alpha \right)}{1-\alpha }x\left(a\right)E_{\alpha }[-\gamma (t{)}^{\alpha }].$(3.1) The right AB Caputo fractional derivatives and the right AB Riemann–Liouville derivative are related by the identity: (3.2) $\begin{array}{r}{}_t^{ABC}{D}_b^{\alpha }x\left(t\right)={}_t^{ABR}{D}_b^{\alpha }x\left(t\right)-\frac{B\left(\alpha \right)}{1-\alpha }x\left(b\right)E_{\alpha }[-\gamma (b-t{)}^{\alpha }].\end{array}$(3.2) There are useful relations between the left and right AB FDs in the Riemann–Liouville and Caputo senses and the associated AB fractional integrals as the following formulas state. (3.3) $\begin{array}{rl}{}_a^{AB}{I}_t^{\alpha }{(}_a^{ABR}{D}_t^{\alpha }x\left)\right(t)& ={}_t^{AB}{I}_b^{\alpha }{(}_t^{ABR}{D}_b^{\alpha }x\left)\right(t)=x(t),\end{array}$(3.3) (3.4) $\begin{array}{rl}{}_a^{AB}{I}_t^{\alpha }{(}_a^{ABC}{D}_t^{\alpha }x\left)\right(t)& =x\left(t\right)-x\left(a\right),\end{array}$(3.4) (3.5) $\begin{array}{rl}{}_t^{AB}{I}_b^{\alpha }{(}_t^{ABC}{D}_T^{\alpha }u\left)\right(t)& =x\left(t\right)-x\left(b\right),\end{array}$(3.5) (3.6) $\begin{array}{rl}{}_t^{ABC}{D}_b^{\alpha }x\left(t\right)& =-{}_0^{ABC}{D}_t^{\alpha }x\left(t\right).\end{array}$(3.6) As a consequence, the backwards in time with the fractional-time derivative with non-singular Mittag–Leffler kernel at the based point T is equivalently written as a forward in time operator with the fractional-time derivative with non-singular Mittag–Leffler kernel $-{}_0^{ABC}{D}_t^{\alpha }.$

Lemma 3.2

[4]

The AB integral operators and ABR differential operators form a commutative family of differintegral operators: (3.7) ${}^{ABR}{D}_{a+}^{\alpha }\left({}^{ABR}{D}_{a+}^{\beta }x\left(t\right)\right){=}^{ABR}{D}_{a+}^{\beta }\left({}^{ABR}{D}_{a+}^{\alpha }x\left(t\right)\right),$(3.7) (3.8) ${}^{AB}{I}_{a+}^{\alpha }\left({}^{AB}{I}_{a+}^{\beta }x\left(t\right)\right){=}^{AB}{I}_{a+}^{\beta }\left({}^{AB}{I}_{a+}^{\alpha }x\left(t\right)\right),$(3.8) (3.9) ${}^{ABR}{D}_{a+}^{\alpha }\left({}^{AB}{I}_{a+}^{\beta }x\left(t\right)\right){=}^{AB}{I}_{a+}^{\beta }\left({}^{ABR}{D}_{a+}^{\alpha }x\left(t\right)\right),$(3.9) for $\alpha ,\beta \in (0,1)$ and a,x satisfying the conditions from Definition (2.1(2.1) $\begin{array}{rl}{}_a^{ABC}{D}_t^{\alpha }x\left(t\right)& =\frac{B\left(\alpha \right)}{1-\alpha }{\int }_a^t\frac{\mathrm{d}}{\mathrm{d}s}x\left(s\right)E_{\alpha }[-\gamma (t-s{)}^{\alpha }]\mathrm{d}s,\\ & \text{(left ABCD)}\end{array}$(2.1) ).

Lemma 3.3

The semigroup property [4]

The semigroup property for AB fractional differintegrals is not satisfied in general. For example, taking $B\left(\alpha \right)=1$ we get ${}^{AB}{I}_{0+}^{2/3}\left(t\right)=\left(\frac{1}{3}+{\frac{2}{3}}^{RL}{I}_{0+}^{2/3}\right)t=\frac{1}{3}t+\frac{2}{3\mathrm{\Gamma }\left(8/3\right)}{t}^{5/3}$ and yet $\begin{array}{rl}{}^{AB}{I}_{0+}^{1/3}\left({}^{AB}{I}_{0+}^{1/3}\left(t\right)\right)& ={}^{AB}{I}_{0+}^{1/3}\left(\frac{2}{3}t+\frac{1}{3\mathrm{\Gamma }\left(7/3\right)}{t}^{4/3}\right)\\ & =\frac{2}{3}\left(\frac{2}{3}t+\frac{1}{3\mathrm{\Gamma }\left(7/3\right)}{t}^{4/3}\right)\\ & +\frac{1}{3\mathrm{\Gamma }\left(7/3\right)}\left(\frac{2}{3}{t}^{4/3}+\frac{\mathrm{\Gamma }\left(7/3\right)}{\mathrm{\Gamma }\left(8/3\right)}{t}^{5/3}\right)\\ & =\frac{4}{9}t+\frac{4}{9\mathrm{\Gamma }\left(7/3\right)}{t}^{4/3}+\frac{1}{3\mathrm{\Gamma }\left(8/3\right)}{t}^{5/3}\end{array}$ – two entirely different expressions.

Can we find conditions for when the semigroup property does hold?

First, note that it will be sufficient to consider fractional integrals only. Any function which satisfies the semigroup property for ABR fractional derivatives generates one which satisfies it for AB fractional integrals and vice versa. This is because (3.10) ${}^{ABR}{I}_{0+}^{\alpha }\left({}^{ABR}{I}_{0+}^{\beta }f\left(t\right)\right)=g\left(t\right){=}^{ABR}{I}_{0+}^{\alpha +\beta }f\left(t\right)$(3.10) is exactly equivalent to (3.11) ${}^{ABR}{D}_{0+}^{\beta }\left({}^{ABR}{D}_{0+}^{\alpha }g\left(t\right)\right)=f\left(t\right){=}^{ABR}{D}_{0+}^{\alpha +\beta }g\left(t\right).$(3.11) This is good to know, because the definition of AB fractional integrals is much simpler and easier to work with than that of ABR fractional derivatives.

Next we state the following proposition which gives the integration by parts.

Lemma 3.4

Integration by parts, see [38]

Let $\alpha >0,p\ge 1,q\ge 1$, and $\left(1/p\right)+\left(1/q\right)\le 1+\alpha (p\ne 1$ and $q\ne 1\text{in the case}\left(1/p\right)+\left(1/q\right)=1+\alpha )$. Then for any $\phi \left(x\right)\in L^p(a,b),\psi \left(x\right)\in L^q(a,b)$, we have (3.12) ${\int }_a^b\phi \left(x\right){}_a^{AB}{I}_t^{\alpha }\psi \left(x\right)\mathrm{d}x={\int }_a^b\psi \left(x\right){}_t^{AB}{I}_b^{\alpha }\phi \left(x\right)\mathrm{d}x,$(3.12) (3.13) ${\int }_a^b\phi \left(x\right){}_t^{AB}{I}_b^{\alpha }\psi \left(x\right)\mathrm{d}x={\int }_a^b\psi \left(x\right){}_a^{AB}{I}_t^{\alpha }\phi \left(x\right)\mathrm{d}x,$(3.13) $\text{if}\phi \left(x\right)\in {}_t^{AB}{I}_b^{\alpha }\left(L^p\right)\text{and}\psi \left(x\right)\in {}_a^{AB}{I}_t^{\alpha }\left(L^q\right),$ then (3.14) $\begin{array}{rl}{\int }_a^b\phi \left(x\right){}_a^{ABR}{D}_t^{\alpha }\psi \left(x\right)\mathrm{d}x& ={\int }_a^b\psi \left(x\right){}_t^{ABR}{D}_b^{\alpha }\phi \left(x\right)\mathrm{d}x,\end{array}$(3.14) (3.15) $\begin{array}{rl}& {\int }_a^b\phi \left(x\right){}_a^{ABC}{D}_t^{\alpha }\psi \left(x\right)\mathrm{d}x\\ & ={\int }_a^b\psi \left(x\right){}_t^{ABR}{D}_b^{\alpha }\phi \left(x\right)\mathrm{d}x\\ & +\frac{B\left(\alpha \right)}{1-\alpha }\psi \left(t\right)E_{\alpha ,1,\left(-\alpha \right)/\left(1-\alpha \right),b}^1\phi \left(t\right){|}_a^b,\end{array}$(3.15) (3.16) $\begin{array}{rl}& {\int }_a^b\phi \left(x\right){}_t^{ABC}{D}_b^{\alpha }\psi \left(x\right)\mathrm{d}x\\ & ={\int }_a^b\psi \left(x\right){}_a^{ABR}{D}_t^{\alpha }\phi \left(x\right)\mathrm{d}x\\ & -\frac{B\left(\alpha \right)}{1-\alpha }\psi \left(t\right)E_{\alpha ,1,\left(-\alpha \right)/\left(1-\alpha \right),a}^1\phi \left(t\right){|}_a^b,\end{array}$(3.16) where the left generalized fractional integral operator (3.17) $\begin{array}{rl}{E}_{\gamma ,\mu ,\omega ,a}^{\alpha }x\left(t\right)& ={\int }_a^t(t-\tau {)}^{\mu -1}{E}_{\gamma ,\mu }^{\alpha }\\ & \times \left[\omega \right(t-\tau {)}^{\gamma }\left]x\right(\tau )\mathrm{d}\tau ,t>a,\end{array}$(3.17) and the right generalized fractional integral operator (3.18) $\begin{array}{rl}{E}_{\gamma ,\mu ,\omega ,b}^{\alpha }x\left(t\right)& ={\int }_t^b(\tau -t{)}^{\mu -1}{E}_{\gamma ,\mu }^{\alpha }\\ & \times \left[\omega \right(\tau -t{)}^{\gamma }\left]x\right(\tau )\mathrm{d}\tau ,t<b.\end{array}$(3.18)

Next we state the following proposition which gives the weak formulation of the problem (1.1(1.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)=f_1\left(t\right),\\ & a.e.t\in ]0,T[,\end{array}$(1.1) )–(1.5(1.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma },$(1.5) ), that will be fundamental in our analysis.

Proposition 3.5

[8]

Let $\phi ,y\in {\mathcal{C}}^{\mathrm{\infty }}\left(\overline{Q}\right)$. Then, we have (3.19) $\begin{array}{rl}& {\int }_{\mathrm{\Omega }}{\int }_0^T\left({}_0^{ABC}{D}_t^{\alpha }\alpha y(x,t)-\mathrm{\Delta }y(x,t)\right)\phi (x,t)\mathrm{d}x\mathrm{d}t\\ & ={\int }_0^T{\int }_{\mathrm{\partial \Omega }}y\frac{\mathrm{\partial }\phi }{\mathrm{\partial }\sigma }\phi \mathrm{d}\sigma \mathrm{d}t-{\int }_0^T{\int }_{\mathrm{\partial \Omega }}\phi \frac{\mathrm{\partial }y}{\mathrm{\partial }\sigma }\mathrm{d}\sigma \mathrm{d}t\\ & +{\int }_{\mathrm{\Omega }}{\int }_0^{T}y(x,t)\left(-{}_t^{ABC}{D}_T\alpha \phi (x,t)-\mathrm{\Delta }\phi (x,t)\right)\mathrm{d}t\mathrm{d}x\\ & -\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^{T}y(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]\phi (x,t)\mathrm{d}t\mathrm{d}x\\ & +\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}\phi (x,T){\int }_0^{T}y(x,t)E_{\alpha ,\alpha }\\ & \times \left[-\gamma (T-t{)}^{\alpha }\right]\mathrm{d}t\mathrm{d}x.\end{array}$(3.19)

We also introduce the space (3.20) $\begin{array}{rl}\mathcal{W}(0,T)& :=\{y:y\in L^2(0,T;H_0^1\left(\mathrm{\Omega }\right)),{}_0^{ABC}{D}_t^{\alpha }y(t)\\ & \in L^2(0,T;H^{-1}(\mathrm{\Omega }\left)\right)\},\end{array}$(3.20) in which a solution of a fractional differential systems is contained. The spaces considered in this paper are assumed to be real.

Lemma 3.6

[2]

Let $0<\alpha <1$, $\mathbb{X}$ be a Banach space and $f\in \mathcal{C}\left(\right[0,T],\mathbb{X})$. Then for all, $t_1,t_2\in [0,T]$ (3.21) ${}_0{I}_t^{\alpha }f\left(t_1\right){-}_0{I}_t^{\alpha }f\left(t_2\right)|{|}_{\mathbb{X}}\le \frac{||f|{|}_{L^{\mathrm{\infty }}\left(\right(0,T);\mathbb{X})}}{\mathrm{\Gamma }(\alpha +1)}|t_1-t_2{|}^{\alpha }.$(3.21)

Remark 3.7

[2]

Since $\mathcal{C}\left(\right[0,T],\mathbb{X})\subset L^{\mathrm{\infty }}\left(\right(0,T);\mathbb{X})\subset L^2\left(\right(0,T);\mathbb{X})$ because $[0,T]$ is a bounded subset of $\mathbb{R}$, Lemma 2.4 holds for $f\in L^2\left(\right(0,T);\mathbb{X})$ and we have that ${}_0{I}_t^{\alpha }f\in \mathcal{C}\left(\right[0,T],\mathbb{X})\subset L^2\left(\right(0,T);\mathbb{X}).$

4. Coupled diffusion system with Atangana–Baleanu derivatives

For $y_{0,1},y_{0,2}\in H_0^1\left(\mathrm{\Omega }\right)$ and $f_1,f_2\in L^2(0,T;H^{-1}(\mathrm{\Omega }\left)\right)$, let us consider the fractional problem for coupled evolution system:

Find $y=\{y_1,y_2\}\in \mathcal{W}(0,T)\times \mathcal{W}(0,T)$ such that (4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) (4.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2\left(y\right)+y_1\left(t\right)\\ & =f_2\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.2) (4.3) $y_1(x,0)=y_{0,1}\left(x\right),x\in \mathrm{\Omega },$(4.3) (4.4) $y_2(x,0)=y_{0,2}\left(x\right),x\in \mathrm{\Omega },$(4.4) (4.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma }.$(4.5) We also need some trace results.

Lemma 4.1

Let $f_1,f_2\in L^2\left(Q\right)$ and $y_1,y_2\in L^2\left(\right(0,T);H_0^1(\mathrm{\Omega }\left)\right)$ be such that ${}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right),{}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)\in L^2\left(Q\right)$ and ${}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)=f_1\left(t\right),{}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2\left(t\right)+y_1\left(t\right)=f_2\left(t\right)$. Then we have

(i) $y_1{|}_{\mathrm{\Sigma }},y_2{|}_{\mathrm{\Sigma }}$ exists and belongs $L^2\left(\right(0,T);H^{-1}(\mathrm{\Gamma }\left)\right)$,

(ii) $y_1\left(0\right),y_2\left(0\right)$ belongs to $L^2\left(\mathrm{\Omega }\right)$.

Proof.

In view of Lemma 2.4, ${}_0^{AB}{I}_t^{\alpha }\left({}_0^{ABC}{D}_t^{\alpha }{y}_1\right(t\left)\right)$, ${}_0^{AB}{I}_t^{\alpha }\left({}_0^{ABC}{D}_t^{\alpha }{y}_2\right(t\left)\right)\in L^2\left(\mathrm{\Omega }\right)$ because ${}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)$, ${}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)\in L^2\left(Q\right)$. Hence, $y_1\left(0\right),y_2\left(0\right)$ exists and belongs to $L^2\left(\mathrm{\Omega }\right)$ since ${}_0^{AB}{I}_t^{\alpha }\left({}_0^{ABC}{D}_t^{\alpha }{y}_1\right(t\left)\right)=y_1\left(t\right)-y_1\left(0\right)$, ${}_0^{AB}{I}_t^{\alpha }\left({}_0^{ABC}{D}_t^{\alpha }{y}_2\right(t\left)\right)=y_2\left(t\right)-y_2\left(0\right)$ and $y_1\left(t\right),y_2\left(t\right)\in L^2\left(\mathrm{\Omega }\right)$.  ▪

For the Laplace operator $\mathrm{\Delta }=\sum _{i=1}^n\left({\mathrm{\partial }}^2/\mathrm{\partial }{x}_i^2\right)$ in (4.1(4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) ), (4.2(4.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2\left(y\right)+y_1\left(t\right)\\ & =f_2\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.2) ), we define the bilinear form $\pi (t;y,\phi )$ as follows.

Definition 4.1

For each $t\in ]0,T[$, $y=(y_1,y_2)$ and $\phi =({\phi }_1,{\phi }_2)$, we define a family of bilinear forms $\pi (t;y,\phi )$ on $\left(H_0^1\right(\mathrm{\Omega }){)}^2$ by (4.6) $\begin{array}{rl}\pi (t;y,\phi )& ={\left(-\mathrm{\Delta }{y}_1+y_1-y_2,{\phi }_1\right)}_{L^2\left(\mathrm{\Omega }\right)}\\ & +{\left(-\mathrm{\Delta }{y}_2+y_2+y_1,{\phi }_2\right)}_{L^2\left(\mathrm{\Omega }\right)},\\ & y,\phi \in \left(H_0^1\right(\mathrm{\Omega }){)}^2,\end{array}$(4.6) which can be written as (4.7) $\begin{array}{rl}\pi (t;y,\phi )& ={\int }_{\mathrm{\Omega }}\left(\mathrm{\nabla }{y}_1\right(x\left){\phi }_1\right(x)+\mathrm{\nabla }{y}_2(x\left){\phi }_2\right(x)\mathrm{d}x\\ & +{\int }_{\mathrm{\Omega }}[y_1{\phi }_1+y_2{\phi }_2-y_2{\phi }_1+y_1{\phi }_2]\mathrm{d}x,\end{array}$(4.7) where $\mathrm{\nabla }=\sum _{i=1}^n\left(\mathrm{\partial }/\mathrm{\partial }{x}_i\right)$ is the grad operator.

Lemma 4.2

The bilinear form $\pi (t;y,\phi )$ in (4.7(4.7) $\begin{array}{rl}\pi (t;y,\phi )& ={\int }_{\mathrm{\Omega }}\left(\mathrm{\nabla }{y}_1\right(x\left){\phi }_1\right(x)+\mathrm{\nabla }{y}_2(x\left){\phi }_2\right(x)\mathrm{d}x\\ & +{\int }_{\mathrm{\Omega }}[y_1{\phi }_1+y_2{\phi }_2-y_2{\phi }_1+y_1{\phi }_2]\mathrm{d}x,\end{array}$(4.7) ) is coercive on $\left(H_0^1\right(\mathrm{\Omega }){)}^2$ that is for $y=(y_1,y_2)$, we have (4.8) $\pi (t;y,y)\ge \lambda ||y|{|}_{\left(H_0^1\right(\mathrm{\Omega }){)}^2}^2,\lambda >0.$(4.8)

Proof.

It is well known that the ellipticity of $\mathrm{\Delta }=\sum _{i=1}^n\left({\mathrm{\partial }}^2/\mathrm{\partial }{x}_i^2\right)$ is sufficient for the coerciveness of $\pi (t;y,\phi )$ on $\left(H_0^1\right(\mathrm{\Omega }){)}^2$. Then we get $\begin{array}{rl}\pi (t;y,y)& ={\int }_{\mathrm{\Omega }}\left(\mathrm{\nabla }{y}_1\right(x\left)y_1\right(x)+\mathrm{\nabla }{y}_2(x\left)y_2\right(x)\mathrm{d}x\\ & +{\int }_{\mathrm{\Omega }}[y_1{y}_1+y_2{y}_2-y_2{y}_1+y_1{y}_2]\mathrm{d}x\\ & \ge \beta \sum _{i=1}^n||\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_i}{y}_1\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2\\ & +\beta ||y_1\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2+||y_1\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2\\ & +\beta \sum _{i=1}^n||\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_i}{y}_2\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2\\ & +\beta ||y_2\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2+||y_2\left(x\right)|{|}_{L^2\left(\mathrm{\Omega }\right)}^2\\ & \ge {\lambda }_1||y_1|{|}_{H_0^1\left(\mathrm{\Omega }\right)}^2+{\lambda }_1||y_2|{|}_{H_0^1\left(\mathrm{\Omega }\right)}^2\\ & \ge \lambda ||y|{|}_{\left(H_0^1\right(\mathrm{\Omega }){)}^2}^2,\lambda =max({\lambda }_1,{\lambda }_2)>0.\end{array}$  ▪

Also we assume that $\mathrm{\forall }y,\phi \in \left(H_0^1\right(\mathrm{\Omega }{)}^2$ the function $t\to \pi (t;y,\phi )$ is continuously differentiable in $]0,T[$ and the bilinear form $\pi (t;y,\phi )$ is symmetric, (4.9) $\pi (t;y,\phi )=\pi (t;\phi ,y)\mathrm{\forall }y,\phi \in \left(H_0^1\right(\mathrm{\Omega }){)}^2.$(4.9) Then (4.1(4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) )–(4.5(4.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma }.$(4.5) ) constitute a fractional Dirichlet coupled problem. First by using the Lax–Milgram lemma, we prove sufficient conditions for the existence of a unique solution of the mixed initial-boundary value problem (4.1(4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) )–(4.5(4.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma }.$(4.5) ).

Lemma 4.3

[23, 24]

(Fractional Green's formula for evolution systems). Let $y=(y_1,y_2)$ be the solution of system (4.1(4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) )–(4.5(4.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma }.$(4.5) ). Then for any $\phi =({\phi }_1,{\phi }_2)\in \left(C^{\mathrm{\infty }}\right(\overline{Q}){)}^2$ such that $\phi (x,T)=({\phi }_1,{\phi }_2)(x,T)=0$ in Ω and $\phi =({\phi }_1,{\phi }_2)=0$ on Σ, we have for each i=1,2 (4.10) $\begin{array}{rl}& {\int }_{\mathrm{\Omega }}{\int }_0^T\left({}_0^{ABC}{D}_t^{\alpha }{y}_i(x,t)-\mathrm{\Delta }{y}_i(x,t)\right){\phi }_i(x,t)\mathrm{d}x\mathrm{d}t\\ & ={\int }_0^T{\int }_{\mathrm{\partial \Omega }}{y}_i\frac{\mathrm{\partial }{\phi }_i}{\mathrm{\partial }\sigma }{\phi }_i\mathrm{d}\sigma \mathrm{d}t-{\int }_0^T{\int }_{\mathrm{\partial \Omega }}{\phi }_i\frac{\mathrm{\partial }{y}_i}{\mathrm{\partial }\sigma }\mathrm{d}\sigma \mathrm{d}t\\ & +{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_i(x,t)\left(-{}_t^{ABC}{D}_T^{\alpha }{\phi }_i(x,t)-\mathrm{\Delta }{\phi }_i(x,t)\right)\mathrm{d}t\mathrm{d}x\\ & -\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_i(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_i(x,t)\mathrm{d}t\mathrm{d}x.\end{array}$(4.10)

Lemma 4.4

If (4.8(4.8) $\pi (t;y,y)\ge \lambda ||y|{|}_{\left(H_0^1\right(\mathrm{\Omega }){)}^2}^2,\lambda >0.$(4.8) ) and (4.9(4.9) $\pi (t;y,\phi )=\pi (t;\phi ,y)\mathrm{\forall }y,\phi \in \left(H_0^1\right(\mathrm{\Omega }){)}^2.$(4.9) ) hold, then the problem (4.1(4.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1\left(t\right)-y_2\left(t\right)\\ & =f_1\left(t\right),a.e.t\in ]0,T[,\end{array}$(4.1) )–(4.5(4.5) $y_1(x,t)=0,y_2(x,t)=0,(x,t)\in \mathrm{\Sigma }.$(4.5) ) admits a unique solution $y\left(t\right)=\left(y_1\right(t),y_2(t\left)\right)\in \left(\mathcal{W}\right(0,T){)}^2$.

Proof.

From the coerciveness condition (4.8(4.8) $\pi (t;y,y)\ge \lambda ||y|{|}_{\left(H_0^1\right(\mathrm{\Omega }){)}^2}^2,\lambda >0.$(4.8) ) and using the Lax–Milgram lemma, there exists a unique element $y\left(t\right)=\left(y_1\right(t),y_2(t\left)\right)\in \left(H_0^1\right(\mathrm{\Omega }){)}^2$ such that (4.11) $\begin{array}{rl}& \left({}_0^{ABC}{D}_t^{\alpha }y\right(t),\phi {)}_{\left(L^2\right(Q){)}^2}+\pi (t;y,\phi )=L(\phi )\text{for all}\\ & \phi =({\phi }_1,{\phi }_2)\in \left(H_0^1\right(\mathrm{\Omega }){)}^2,\end{array}$(4.11) where $L\left(\phi \right)$ is a continuous linear form on $\left(H_0^1\right(\mathrm{\Omega }){)}^2$ and takes the form (4.12) $\begin{array}{rl}L\left(\phi \right)& ={\int }_Q[f_1{\phi }_1+f_2{\phi }_2]\mathrm{d}x\mathrm{d}t-\frac{B\left(\alpha \right)}{1-\alpha }\\ & \times {\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,1}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_1(x,t)\mathrm{d}t\mathrm{d}x-\frac{B\left(\alpha \right)}{1-\alpha }\\ & \times {\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,2}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_2(x,t)\mathrm{d}t\mathrm{d}x,\\ f& =(f_1,f_2)\in \left(L^2\right(Q){)}^2,y_0=(y_{0,1},y_{0,2})\in (L^2\left(\mathrm{\Omega }\right){)}^2.\end{array}$(4.12) Then Equation (4.11(4.11) $\begin{array}{rl}& \left({}_0^{ABC}{D}_t^{\alpha }y\right(t),\phi {)}_{\left(L^2\right(Q){)}^2}+\pi (t;y,\phi )=L(\phi )\text{for all}\\ & \phi =({\phi }_1,{\phi }_2)\in \left(H_0^1\right(\mathrm{\Omega }){)}^2,\end{array}$(4.11) ) equivalents to there exists a unique solution $y\left(t\right)=\left(y_1\right(t),y_2(t\left)\right)\in \left(H_0^1\right(\mathrm{\Omega }){)}^2$ for (4.13) $\begin{array}{rl}& {\left({}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1+y_2,{\phi }_1\left(x\right)\right)}_{L^2\left(Q\right)}\\ & +{\left({}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2-y_1,{\phi }_2\left(x\right)\right)}_{L^2\left(Q\right)}\\ & =L\left(\phi \right).\end{array}$(4.13) Then Equation (4.13(4.13) $\begin{array}{rl}& {\left({}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1+y_2,{\phi }_1\left(x\right)\right)}_{L^2\left(Q\right)}\\ & +{\left({}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2-y_1,{\phi }_2\left(x\right)\right)}_{L^2\left(Q\right)}\\ & =L\left(\phi \right).\end{array}$(4.13) ) is equivalent to the fractional evolution equations (4.14) ${}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1+y_2=f_1,$(4.14) (4.15) ${}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2-y_1=f_2,$(4.15) “tested” against ${\phi }_1\left(x\right),{\phi }_2\left(x\right)$ respectively.

Let us multiply both sides in (4.14(4.14) ${}_0^{ABC}{D}_t^{\alpha }{y}_1\left(t\right)-\mathrm{\Delta }{y}_1\left(t\right)+y_1+y_2=f_1,$(4.14) ), (4.15(4.15) ${}_0^{ABC}{D}_t^{\alpha }{y}_2\left(t\right)-\mathrm{\Delta }{y}_2\left(t\right)+y_2-y_1=f_2,$(4.15) ) by ${\phi }_1\left(x\right)$, ${\phi }_2\left(x\right)$ respectively and applying Green's formula (Lemma 3.6), we have (4.16) $\begin{array}{rl}& {\int }_Q\left({}_0^{ABC}{D}_t^{\alpha }{y}_1\right(t)-\mathrm{\Delta }{y}_1(t)+y_1+y_2){\phi }_1\left(x\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q{f}_1{\phi }_1\mathrm{d}x\mathrm{d}t\text{for all}{\phi }_1\left(x\right)\in H_0^1\left(\mathrm{\Omega }\right),\end{array}$(4.16) (4.17) $\begin{array}{rl}& {\int }_Q\left({}_0^{ABC}{D}_t^{\alpha }{y}_2\right(t)-\mathrm{\Delta }{y}_2(t)+y_2-y_1){\phi }_2\left(x\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q{f}_2{\phi }_2\mathrm{d}x\mathrm{d}t\text{for all}{\phi }_2\left(x\right)\in H_0^1\left(\mathrm{\Omega }\right)\end{array}$(4.17) applying Green's formula (Corollary 3.6), we have $\begin{array}{rl}& -\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_1(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_1(x,t)\mathrm{d}t\mathrm{d}x\\ & +{\int }_0^T{\int }_{\mathrm{\partial \Omega }}{y}_1\frac{\mathrm{\partial }{\phi }_1}{p\nu }\mathrm{d}\mathrm{\Gamma }\mathrm{d}t-{\int }_0^T{\int }_{\mathrm{\partial \Omega }}\mathrm{\partial }{y}_{1}p\nu \phi \mathrm{d}\mathrm{\Gamma }\mathrm{d}t\\ & +{\int }_0^T{\int }_{\mathrm{\Omega }}{y}_1(x,t)(-{}_0^{ABC}{D}_t^{\alpha }{\phi }_1(x,t)+\mathrm{\Delta }{\phi }_1(x,t\left)\right)\mathrm{d}x\mathrm{d}t\\ & +{\int }_Q(y_1+y_2){\phi }_1\mathrm{d}x\mathrm{d}t={\int }_Q{f}_1{\phi }_1\left(x\right)\mathrm{d}x\mathrm{d}t\\ & -\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_2(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_2(x,t)\mathrm{d}t\mathrm{d}x\\ & +{\int }_0^T{\int }_{\mathrm{\partial \Omega }}{y}_2\frac{\mathrm{\partial }{\phi }_1}{p\nu }\mathrm{d}\mathrm{\Gamma }\mathrm{d}t-{\int }_0^T{\int }_{\mathrm{\partial \Omega }}\mathrm{\partial }{y}_{2}p\nu \phi \mathrm{d}\mathrm{\Gamma }\mathrm{d}t\\ & +{\int }_0^T{\int }_{\mathrm{\Omega }}{y}_2(x,t)(-{}_0^{ABC}{D}_t^{\alpha }{\phi }_2(x,t)+\mathrm{\Delta }{\phi }_2(x,t\left)\right)\mathrm{d}x\mathrm{d}t\\ & +{\int }_Q(y_2-y_1){\phi }_2\mathrm{d}x\mathrm{d}t={\int }_Q{f}_2{\phi }_2\left(x\right)\mathrm{d}x\mathrm{d}t\end{array}$ whence comparing with (4.11(4.11) $\begin{array}{rl}& \left({}_0^{ABC}{D}_t^{\alpha }y\right(t),\phi {)}_{\left(L^2\right(Q){)}^2}+\pi (t;y,\phi )=L(\phi )\text{for all}\\ & \phi =({\phi }_1,{\phi }_2)\in \left(H_0^1\right(\mathrm{\Omega }){)}^2,\end{array}$(4.11) ), (4.12(4.12) $\begin{array}{rl}L\left(\phi \right)& ={\int }_Q[f_1{\phi }_1+f_2{\phi }_2]\mathrm{d}x\mathrm{d}t-\frac{B\left(\alpha \right)}{1-\alpha }\\ & \times {\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,1}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_1(x,t)\mathrm{d}t\mathrm{d}x-\frac{B\left(\alpha \right)}{1-\alpha }\\ & \times {\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,2}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_2(x,t)\mathrm{d}t\mathrm{d}x,\\ f& =(f_1,f_2)\in \left(L^2\right(Q){)}^2,y_0=(y_{0,1},y_{0,2})\in (L^2\left(\mathrm{\Omega }\right){)}^2.\end{array}$(4.12) ), we get $\begin{array}{rl}& \frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_1(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_1(x,t)\mathrm{d}t\mathrm{d}x\\ & -{\int }_0^T{\int }_{\mathrm{\partial \Omega }}{y}_1\frac{\mathrm{\partial }{\phi }_1}{p\nu }\mathrm{d}\mathrm{\Gamma }\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,1}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_1(x,t)\mathrm{d}t\mathrm{d}x,\\ & \frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_2(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_2(x,t)\mathrm{d}t\mathrm{d}x\\ & -{\int }_0^T{\int }_{\mathrm{\partial \Omega }}{y}_2\frac{\mathrm{\partial }{\phi }_2}{p\nu }\mathrm{d}\mathrm{\Gamma }\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{y}_{0,2}{E}_{\alpha ,\alpha }[-\gamma t^{\alpha }]{\phi }_2(x,t)\mathrm{d}t\mathrm{d}x.\end{array}$ From this we deduce the initial conditions $\begin{array}{rl}{y}_1(x,0)& =y_{0,1},x\in \mathrm{\Omega },\\ y_2(x,0)& =y_{0,2},x\in \mathrm{\Omega },\end{array}$ which completes the proof.  ▪

5. Optimization theorem and the fractional control problem

For a control $u=(u_1,u_2)\in \left(L^2\right(Q){)}^2$, the state $y\left(u\right)=\left(y_1\right(u),y_1(u\left)\right)$ of the system is given by the fractional variation coupled systems: (5.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right)+u_1,\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(5.1) (5.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(u\right)-\mathrm{\Delta }{y}_2\left(u\right)+y_2\left(u\right)+y_1\left(u\right)\\ & =f_2\left(t\right)+u_2,\text{in}Q,a.e.t\in ]0,T[,f_2\in L^2\left(Q\right),\end{array}$(5.2) (5.3) $y_1(x,0;u)=y_{0,1}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(5.3) (5.4) $y_2(x,0;u)=y_{0,2}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(5.4) (5.5) $y_1(x,t)=0,y_2(x,t)=0,x\in \mathrm{\Gamma },t\in (0,T).$(5.5) The final observation equations are given by (5.6) $z_i\left(u\right)=y_i(u,T),\text{for each}i=1,2.$(5.6) The cost function $J\left(v\right)$ for $v=\{v_1,v_2\}$ is given by (5.7) $\begin{array}{rl}J\left(v\right)& ={\int }_Q\left[\right(y_1(v,T)-z_{d,1}{)}^2+\left(y_2\right(v,T)-z_{d,2}{)}^2]\mathrm{d}x\mathrm{d}t\\ & +(Nv,v{)}_{\left(L^2\right(Q){)}^2},\end{array}$(5.7) where $z_d=\{z_{d,1},z_{d,2}\}$ is a given element in $\left(L^2\right(Q){)}^2$ and $N=\{N_1,N_2\}\in \mathcal{L}\left(L^2\right(Q),L^2(Q\left)\right)$ is Hermitian positive definite operator: (5.8) $(N{u}_i,u_i)\ge c||u_i|{|}_{L^2\left(Q\right)}^2,c>0,\text{for each}i=1,2.$(5.8) Control constraints: We define $U_{ad}$( set of admissible controls) is closed, convex subset of $U=L^2\left(Q\right)\times L^2\left(Q\right)$.

Control problem: We want to minimize J over $U_{ad}$, i.e. find $u=\{u_1,u_2\}$ such that (5.9) $J\left(u\right)=\underset{v=\{v_1,v_2\}\in U_{ad}}{inf}J\left(v\right).$(5.9) Under the given considerations, we have the following theorem.

Theorem 5.1

The problem (5.9(5.9) $J\left(u\right)=\underset{v=\{v_1,v_2\}\in U_{ad}}{inf}J\left(v\right).$(5.9) ) admits a unique solution given by (5.1(5.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right)+u_1,\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(5.1) )–(5.5(5.5) $y_1(x,t)=0,y_2(x,t)=0,x\in \mathrm{\Gamma },t\in (0,T).$(5.5) ) and the optimality condition (5.10) $\begin{array}{rl}& {\int }_Q\left[p_1\right(v_1-u_1)+p_2(v_2-u_2\left)\right]\mathrm{d}x\mathrm{d}t+(Nu,v-u{)}_U\\ & \ge 0,,\mathrm{\forall }v\in U_{ad},u\in U_{ad},\end{array}$(5.10) where $p\left(u\right)=\left\{p_1\right(u),p_2(u\left)\right\}$ is the adjoint state.

Proof.

Since the control $u\in U_{ad}$ is optimal if and only if (5.11) $J^{\mathrm{\prime }}\left(u\right)(v-u)\ge 0\text{for all}v\in U_{ad}.$(5.11) The above condition, when explicitly calculated for this case, gives (5.12) $\begin{array}{rl}& \left(y_1\right(u)-z_{d,1},y_1(v)-y_1(u){)}_{L^2\left(Q\right)}+(y_2\left(u\right)-z_{d,2},y_2\left(v\right)\\ & -y_2\left(u\right){)}_{L^2\left(Q\right)}+(Nu,v-u{)}_U\ge 0,\end{array}$(5.12) i.e. (5.13) $\begin{array}{rl}& {\int }_Q\left[y_1\right(u)-z_{d,1})\left(y_1\right(v)-y_1(u\left)\right)+\left(y_2\right(u)-z_{d,2})\left(y_2\right(v)\\ & -y_2\left(u\right)\left)\right]\mathrm{d}x\mathrm{d}t+(Nu,v-u{)}_{\left(L^2\right(Q){)}^2}\ge 0.\end{array}$(5.13) For the control $u\in \left(L^2\right(Q){)}^2$, the adjoint state $p\left(u\right)=\left\{p_1\right(u),p_2(u\left)\right\}\in \left(L^2\right(Q){)}^2$ is defined by (5.14) $\begin{array}{rl}& -{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)\\ & =y_1\left(u\right)-z_{d,1},\text{in}Q,\end{array}$(5.14) (5.15) $\begin{array}{rl}& -{}_t^{ABC}{D}_T^{\alpha }{p}_2\left(u\right)+\mathrm{\Delta }{p}_2\left(u\right)+p_2\left(u\right)-p_1\left(u\right)\\ & =y_2\left(u\right)-z_{d,2},\text{in}Q,\end{array}$(5.15) (5.16) $p_1\left(u\right)=0,p_2\left(u\right)=0\text{on}\mathrm{\Sigma },$(5.16) (5.17) $p_1(x,T;u)=0,p_2(x,T;u)=0\text{in}\mathrm{\Omega }.$(5.17) Now, multiplying Equations (5.14(5.14) $\begin{array}{rl}& -{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)\\ & =y_1\left(u\right)-z_{d,1},\text{in}Q,\end{array}$(5.14) ) and (5.15(5.15) $\begin{array}{rl}& -{}_t^{ABC}{D}_T^{\alpha }{p}_2\left(u\right)+\mathrm{\Delta }{p}_2\left(u\right)+p_2\left(u\right)-p_1\left(u\right)\\ & =y_2\left(u\right)-z_{d,2},\text{in}Q,\end{array}$(5.15) ) by $\left(y_1\right(v\left)-y_1\right(u\left)\right)$, $\left(y_2\right(v\left)-y_2\right(u\left)\right)$ respectively and applying Green's formula, we obtain (5.18) $\begin{array}{rl}& {\int }_Q\left(y_1\right(u)-z_{d,1})\left(y_1\right(v)-y_1(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q[-{}_t^{ABC}{D}_T^{\alpha }{p}_1(u)+\mathrm{\Delta }{p}_1(u)+p_1(u)\\ & +p_2\left(u\right)\left]\right(y_1\left(v\right)-y_1\left(u\right))\mathrm{d}x\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{p}_1(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]\left(y_1\right(v;x,0)\\ & -y_1(u;x,0))\mathrm{d}t\mathrm{d}x\\ & +{\int }_{\mathrm{\Sigma }}{p}_1\left(u\right)(\frac{\mathrm{\partial }{y}_1\left(v\right)}{\mathrm{\partial }\nu }-\frac{\mathrm{\partial }{y}_1\left(u\right)}{\mathrm{\partial }{\nu }_{\mathcal{A}}})\mathrm{d}\mathrm{\Sigma }\\ & -{\int }_{\mathrm{\Sigma }}\frac{\mathrm{\partial }{p}_1\left(u\right)}{\mathrm{\partial }\nu }\left(y_1\right(v)-y_1(u\left)\right)\mathrm{d}\mathrm{\Sigma }\\ & +{\int }_Q\left[p_1\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)\\ & +p_2\left(u\right)\left]\right(y_1\left(v\right)-y_1\left(u\right))\mathrm{d}x\mathrm{d}t,\end{array}$(5.18) (5.19) $\begin{array}{rl}& {\int }_Q\left(y_2\right(u)-z_{d,2})\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q[-{}_t^{ABC}{D}_T^{\alpha }{p}_2(u)+\mathrm{\Delta }{p}_2(u)+p_2(u)\\ & -p_1\left(u\right)\left]\right(y_2\left(v\right)-y_2\left(u\right))\mathrm{d}x\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{p}_2(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]\left(y_2\right(v;x,0)\\ & -y_2(u;x,0))\mathrm{d}t\mathrm{d}x\\ & +{\int }_{\mathrm{\Sigma }}{p}_2\left(u\right)(\frac{\mathrm{\partial }{y}_2\left(v\right)}{\mathrm{\partial }\nu }-\frac{\mathrm{\partial }{y}_2\left(u\right)}{\mathrm{\partial }\nu })\mathrm{d}\mathrm{\Sigma }\\ & -{\int }_{\mathrm{\Sigma }}\frac{\mathrm{\partial }{p}_2\left(u\right)}{\mathrm{\partial }\nu }\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}\mathrm{\Sigma }\\ & +{\int }_Q\left[p_2\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)\\ & -p_1\left(u\right)\left)\right]\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}x\mathrm{d}t.\end{array}$(5.19) Using (5.1(5.1) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right)+u_1,\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(5.1) )–(5.5(5.5) $y_1(x,t)=0,y_2(x,t)=0,x\in \mathrm{\Gamma },t\in (0,T).$(5.5) ), (5.16(5.16) $p_1\left(u\right)=0,p_2\left(u\right)=0\text{on}\mathrm{\Sigma },$(5.16) ) and (5.17(5.17) $p_1(x,T;u)=0,p_2(x,T;u)=0\text{in}\mathrm{\Omega }.$(5.17) ), we have (5.20) $\begin{array}{rl}& {\int }_Q\left[p_1\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)+p_2(u\left)\right]\left(y_1\right(v)\\ & -y_1\left(u\right))\mathrm{d}x\mathrm{d}t={\int }_Q{p}_1(u\left)\right(v_1-u_1)\mathrm{d}x\mathrm{d}t,\end{array}$(5.20) (5.21) $\begin{array}{rl}& {\int }_Q\left[p_2\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)-p_1(u\left)\right)\left]\right(y_2\left(v\right)\\ & -y_2\left(u\right))\mathrm{d}x\mathrm{d}t={\int }_Q{p}_2(u\left)\right(v_2-u_2)\mathrm{d}x\mathrm{d}t,\end{array}$(5.21) (5.22) $y_1\left(u\right){|}_{\mathrm{\Sigma }}=0,y_2\left(u\right){|}_{\mathrm{\Sigma }}=0p_1\left(u\right){|}_{\mathrm{\Sigma }}=0,p_2\left(u\right){|}_{\mathrm{\Sigma }}=0,$(5.22) (5.23) $y_1(v;x,0)-y_1(u;x,0)=y_{0,1}\left(x\right)-y_{0,1}\left(x\right)=0,$(5.23) (5.24) $y_2(v;x,0)-y_2(u;x,0)=y_{0,2}\left(x\right)-y_{0,2}\left(x\right)=0.$(5.24) Then (5.18(5.18) $\begin{array}{rl}& {\int }_Q\left(y_1\right(u)-z_{d,1})\left(y_1\right(v)-y_1(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q[-{}_t^{ABC}{D}_T^{\alpha }{p}_1(u)+\mathrm{\Delta }{p}_1(u)+p_1(u)\\ & +p_2\left(u\right)\left]\right(y_1\left(v\right)-y_1\left(u\right))\mathrm{d}x\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{p}_1(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]\left(y_1\right(v;x,0)\\ & -y_1(u;x,0))\mathrm{d}t\mathrm{d}x\\ & +{\int }_{\mathrm{\Sigma }}{p}_1\left(u\right)(\frac{\mathrm{\partial }{y}_1\left(v\right)}{\mathrm{\partial }\nu }-\frac{\mathrm{\partial }{y}_1\left(u\right)}{\mathrm{\partial }{\nu }_{\mathcal{A}}})\mathrm{d}\mathrm{\Sigma }\\ & -{\int }_{\mathrm{\Sigma }}\frac{\mathrm{\partial }{p}_1\left(u\right)}{\mathrm{\partial }\nu }\left(y_1\right(v)-y_1(u\left)\right)\mathrm{d}\mathrm{\Sigma }\\ & +{\int }_Q\left[p_1\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)\\ & +p_2\left(u\right)\left]\right(y_1\left(v\right)-y_1\left(u\right))\mathrm{d}x\mathrm{d}t,\end{array}$(5.18) ) becomes (5.25) $\begin{array}{rl}& {\int }_Q\left(y_1\right(u)-z_{d,1})\left(y_1\right(v)-y_1(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q{p}_1\left(u\right)(v_1-u_1)\mathrm{d}x\mathrm{d}t,\end{array}$(5.25) and (5.19(5.19) $\begin{array}{rl}& {\int }_Q\left(y_2\right(u)-z_{d,2})\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q[-{}_t^{ABC}{D}_T^{\alpha }{p}_2(u)+\mathrm{\Delta }{p}_2(u)+p_2(u)\\ & -p_1\left(u\right)\left]\right(y_2\left(v\right)-y_2\left(u\right))\mathrm{d}x\mathrm{d}t\\ & =-\frac{B\left(\alpha \right)}{1-\alpha }{\int }_{\mathrm{\Omega }}{\int }_0^T{p}_2(x,0)E_{\alpha ,\alpha }[-\gamma t^{\alpha }]\left(y_2\right(v;x,0)\\ & -y_2(u;x,0))\mathrm{d}t\mathrm{d}x\\ & +{\int }_{\mathrm{\Sigma }}{p}_2\left(u\right)(\frac{\mathrm{\partial }{y}_2\left(v\right)}{\mathrm{\partial }\nu }-\frac{\mathrm{\partial }{y}_2\left(u\right)}{\mathrm{\partial }\nu })\mathrm{d}\mathrm{\Sigma }\\ & -{\int }_{\mathrm{\Sigma }}\frac{\mathrm{\partial }{p}_2\left(u\right)}{\mathrm{\partial }\nu }\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}\mathrm{\Sigma }\\ & +{\int }_Q\left[p_2\right(u\left)\right({}_0^{ABC}{D}_t^{\alpha }+\mathrm{\Delta }+1)\\ & -p_1\left(u\right)\left)\right]\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}x\mathrm{d}t.\end{array}$(5.19) ) becomes (5.26) $\begin{array}{rl}& {\int }_Q\left(y_2\right(u)-z_{d,2})\left(y_2\right(v)-y_2(u\left)\right)\mathrm{d}x\mathrm{d}t\\ & ={\int }_Q{p}_2\left(u\right)(v_2-u_2)\mathrm{d}x\mathrm{d}t,\end{array}$(5.26) and hence (5.13(5.13) $\begin{array}{rl}& {\int }_Q\left[y_1\right(u)-z_{d,1})\left(y_1\right(v)-y_1(u\left)\right)+\left(y_2\right(u)-z_{d,2})\left(y_2\right(v)\\ & -y_2\left(u\right)\left)\right]\mathrm{d}x\mathrm{d}t+(Nu,v-u{)}_{\left(L^2\right(Q){)}^2}\ge 0.\end{array}$(5.13) ) is equivalent to (5.27) $\begin{array}{rl}& {\int }_Q{p}_1\left(u\right)(v_1-u_1)\mathrm{d}x\mathrm{d}t+{\int }_Q{p}_2\left(u\right)(v_2-u_2)\mathrm{d}x\mathrm{d}t\\ & +(Nu,v-u{)}_{\left(L^2\right(Q){)}^2}\ge 0,\end{array}$(5.27) which can be written as (5.28) $\begin{array}{rl}& {\int }_Q\left[\right(p_1\left(u\right)+N_1{u}_1\left)\right(v_1-u_1)\\ & +\left(p_2\right(u)+N_2{u}_2)(v_2-u_2)\mathrm{d}x\mathrm{d}t\ge 0,\end{array}$(5.28) which completes the proof.  ▪

6. Mathematical examples and applications

This section is devoted to study some mathematical examples and applications to illustrate our results in this paper.

Example 6.1

Neumann problem

We consider an example of a diffusion equation which is analogous to that considered in Section 2 but with Neumann boundary condition and boundary control.

In this example, we consider the space (6.1) $\begin{array}{rl}\mathcal{W}(0,T)& :=\{y:y\in L^2(0,T;H^1\left(\mathrm{\Omega }\right)),{}_0^{ABC}{D}_t^{\alpha }y(t)\\ & \in L^2(0,T;(H^1\left(\mathrm{\Omega }\right){)}^{\mathrm{\prime }}\left)\right\}\end{array}$(6.1) in which a solution of a fractional differential systems is contained. Let $y\left(u\right)=\left\{y_1\right(u),y_2(u\left)\right\}\in \mathcal{W}(0,T)$ be the state of the system which is given by (6.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.2) (6.3) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(u\right)-\mathrm{\Delta }{y}_2\left(u\right)+y_2\left(u\right)+y_1\left(u\right)\\ & =f_2\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_2\in L^2\left(Q\right),\end{array}$(6.3) (6.4) $y_1(x,0;u)=y_{0,1}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(6.4) (6.5) $y_2(x,0;u)=y_{0,2}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(6.5) (6.6) $\frac{\mathrm{\partial }{y}_1(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_1,\frac{\mathrm{\partial }{y}_2(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_2,x\in \mathrm{\Gamma },t\in (0,T).$(6.6) The control $u=\{u_1,u_2\}$ is taken in $L^2\left(\mathrm{\Sigma }\right)\times L^2\left(\mathrm{\Sigma }\right)$: (6.7) $u=\{u_1,u_2\}\in U=L^2\left(\mathrm{\Sigma }\right)\times L^2\left(\mathrm{\Sigma }\right).$(6.7) Problem (6.2(6.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.2) )–(6.6(6.6) $\frac{\mathrm{\partial }{y}_1(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_1,\frac{\mathrm{\partial }{y}_2(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_2,x\in \mathrm{\Gamma },t\in (0,T).$(6.6) ) admits a unique solution. To see this we apply Theorem (1.2) [15], with (6.8) $V=H^1\left(\mathrm{\Omega }\right)\times H^1\left(\mathrm{\Omega }\right),\phi =\{{\phi }_1,{\phi }_2\}\in V,$(6.8) (6.9) $\begin{array}{rl}\pi (t;y,\phi )& =\pi (y,\phi )={\int }_{\mathrm{\Omega }}\mathrm{\Delta }{y}_1\left(x\right){\phi }_1\left(x\right)\mathrm{d}x\\ & +{\int }_{\mathrm{\Omega }}\mathrm{\Delta }{y}_2\left(x\right){\phi }_2\left(x\right)\mathrm{d}x\\ & +{\int }_{\mathrm{\Omega }}[y_1{\phi }_1+y_2{\phi }_2-y_2{\phi }_1+y_1{\phi }_2]\mathrm{d}x,\end{array}$(6.9) (6.10) $\begin{array}{rl}L\left(\phi \right)& =(f,\phi )={\int }_{\mathrm{\Omega }}\left[f_1\right(x,t\left){\phi }_1\right(x)+f_2(x,t\left){\phi }_2\right(x\left)\right]\mathrm{d}x\\ & +{\int }_{\mathrm{\Gamma }}\left[u_1\right(t\left){\phi }_1\right(x)+u_2(t\left){\phi }_2\right(x\left)\right]\mathrm{d}\mathrm{\Gamma }.\end{array}$(6.10) Let us consider the case where we have partial observation of the final state (6.11) $z\left(v\right)=y_1(x,T;v),$(6.11) and the cost function $J\left(v\right)$ for $v=\{v_1,v_2\}$ is given by (6.12) $\begin{array}{rl}J\left(v\right)& ={\int }_{\mathrm{\Omega }}\left(y_1\right(x,T;v)-z_d{)}^2\mathrm{d}x\\ & +(Nv,v{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^2},z_d\in L^2(\mathrm{\Omega }),\end{array}$(6.12) where $N=\{N_1,N_2\}\in \mathcal{L}\left(L^2\right(\mathrm{\Sigma }),L^2(\mathrm{\Sigma }\left)\right)$ is Hermitian positive definite operator: (6.13) $(Nu,u)\ge c||u|{|}_{L^2\left(\mathrm{\Sigma }\right)}^2,c>0.$(6.13) Control constraints: We define $U_{ad}$ (set of admissible controls) is closed, convex subset of $U=L^2\left(\mathrm{\Sigma }\right)\times L^2\left(\mathrm{\Sigma }\right)$.

Control problem: We want to minimize J over $U_{ad}$, i.e. find $u=\{u_1,u_2\}$ such that (6.14) $J\left(u\right)=\underset{v=\{v_1,v_2\}\in U_{ad}}{inf}J\left(v\right).$(6.14) The adjoint state is given by (6.15) $-{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)=0,\text{in}Q,$(6.15) (6.16) $-{}_t^{ABC}{D}_T^{\alpha }{p}_2\left(u\right)+\mathrm{\Delta }{p}_2\left(u\right)+p_2\left(u\right)-p_1\left(u\right)=0,\text{in}Q,$(6.16) (6.17) $\frac{\mathrm{\partial }{p}_1\left(u\right)}{\mathrm{\partial }\nu }=0,\frac{\mathrm{\partial }{p}_2\left(u\right)}{\mathrm{\partial }\nu }=0,\text{on}\mathrm{\Sigma },$(6.17) (6.18) $\begin{array}{rl}{p}_1(x,T;u)& =y_1\left(u\right)-z_d,\text{in}\mathrm{\Omega },\\ p_2(x,T;u)& =0,\text{in}\mathrm{\Omega }.\end{array}$(6.18) The optimality condition is (6.19) $\begin{array}{rl}& {\int }_{\mathrm{\Sigma }}\left[p_1\right(u\left)\right(v_1-u_1)+p_2(u\left)\right(v_2-u_2\left)\right]\mathrm{d}\mathrm{\Sigma }\\ & +(Nu,v-u{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^2}\ge 0,\mathrm{\forall }v\in U_{ad},u\in U_{ad}.\end{array}$(6.19)

Example 6.2

No constraints problem

In the case of no constraint on the control $(U=U_{ad})$ and $N=\{N_1,N_2\}$ is a diagonal matrix of operators. Then (6.19(6.19) $\begin{array}{rl}& {\int }_{\mathrm{\Sigma }}\left[p_1\right(u\left)\right(v_1-u_1)+p_2(u\left)\right(v_2-u_2\left)\right]\mathrm{d}\mathrm{\Sigma }\\ & +(Nu,v-u{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^2}\ge 0,\mathrm{\forall }v\in U_{ad},u\in U_{ad}.\end{array}$(6.19) ) reduces to (6.20) $p_1+N_1{u}_1=0,\text{on}\mathrm{\Sigma },p_2+N_2{u}_2=0,\text{on}\mathrm{\Sigma },$(6.20) equivalent to (6.21) $u_1=-N_1^{-1}\left(p_1\right(u\left){|}_{\mathrm{\Sigma })}\right),u_2=-N_2^{-1}\left(p_2\right(u\left){|}_{\mathrm{\Sigma }}\right).$(6.21) The fractional control is obtained by solving (6.2(6.2) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.2) )–(6.6(6.6) $\frac{\mathrm{\partial }{y}_1(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_1,\frac{\mathrm{\partial }{y}_2(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_2,x\in \mathrm{\Gamma },t\in (0,T).$(6.6) ) and (6.15(6.15) $-{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)=0,\text{in}Q,$(6.15) )–(6.18(6.18) $\begin{array}{rl}{p}_1(x,T;u)& =y_1\left(u\right)-z_d,\text{in}\mathrm{\Omega },\\ p_2(x,T;u)& =0,\text{in}\mathrm{\Omega }.\end{array}$(6.18) ) simultaneous, (where we eliminate $u_1,u_2$ with the aid of (6.21(6.21) $u_1=-N_1^{-1}\left(p_1\right(u\left){|}_{\mathrm{\Sigma })}\right),u_2=-N_2^{-1}\left(p_2\right(u\left){|}_{\mathrm{\Sigma }}\right).$(6.21) )) and then utilizing (6.21(6.21) $u_1=-N_1^{-1}\left(p_1\right(u\left){|}_{\mathrm{\Sigma })}\right),u_2=-N_2^{-1}\left(p_2\right(u\left){|}_{\mathrm{\Sigma }}\right).$(6.21) ).

Example 6.3

If we take (6.22) $\begin{array}{rl}{\mathcal{U}}_{ad}& =\left\{u_i|u_i\in L^2\left(\mathrm{\Sigma }\right),u_i\ge 0\right.\\ & \left.\text{almost everywhere on}\mathrm{\Sigma },i=1,2\right\},\end{array}$(6.22) and $N=\nu \times Identity$, (6.19(6.19) $\begin{array}{rl}& {\int }_{\mathrm{\Sigma }}\left[p_1\right(u\left)\right(v_1-u_1)+p_2(u\left)\right(v_2-u_2\left)\right]\mathrm{d}\mathrm{\Sigma }\\ & +(Nu,v-u{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^2}\ge 0,\mathrm{\forall }v\in U_{ad},u\in U_{ad}.\end{array}$(6.19) ) gives (6.23) $\begin{array}{rl}{u}_1\ge 0,p_1\left(u\right)+{\nu }_1{u}_1& \ge 0,\\ u_1\left(p_1\right(u)+{\nu }_1{u}_1)& =0,\text{on}\mathrm{\Sigma },\end{array}$(6.23) (6.24) $\begin{array}{rl}{u}_2\ge 0,p_2\left(u\right)+{\nu }_2{u}_2& \ge 0,\\ u_2\left(p_2\right(u)+{\nu }_2{u}_2)& =0,\text{on}\mathrm{\Sigma }.\end{array}$(6.24) The fractional optimal control is obtained by the solution of the fractional problem (6.25) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\\ & {}_0^{ABC}{D}_t^{\alpha }{y}_2\left(u\right)-\mathrm{\Delta }{y}_2\left(u\right)+y_2\left(u\right)+y_1\left(u\right)\\ & =f_2\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_2\in L^2\left(Q\right),\end{array}$(6.25) (6.26) $\begin{array}{rl}& -{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)=0,\text{in}Q,\\ & -{}_t^{ABC}{D}_T^{\alpha }{p}_2\left(u\right)+\mathrm{\Delta }{p}_2\left(u\right)+p_2\left(u\right)-p_1\left(u\right)=0,\text{in}Q,\end{array}$(6.26) (6.27) $\begin{array}{rl}{y}_1(x,0;u)& =y_{0,1}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },\\ y_2(x,0;u)& =y_{0,2}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },\end{array}$(6.27) (6.28) $\begin{array}{rl}{p}_1(x,T;u)& =y_1\left(u\right)-z_d,\text{in}\mathrm{\Omega },\\ p_2(x,T;u)& =0,\text{in}\mathrm{\Omega },\end{array}$(6.28) (6.29) $\begin{array}{rl}& \frac{\mathrm{\partial }{y}_1(x,t)}{\mathrm{\partial }\nu }{|}_{\mathrm{\Sigma }}\ge 0,x\in \mathrm{\Gamma },t\in (0,T),\\ & \frac{\mathrm{\partial }{y}_2(x,t)}{\mathrm{\partial }\nu }{|}_{\mathrm{\Sigma }}\ge 0,x\in \mathrm{\Gamma },t\in (0,T),\end{array}$(6.29) (6.30) $\begin{array}{rl}& \begin{array}{rl}& \frac{\mathrm{\partial }{p}_1\left(u\right)}{\mathrm{\partial }\nu }=0,\text{on}\mathrm{\Sigma },\\ & \frac{\mathrm{\partial }{p}_2\left(u\right)}{\mathrm{\partial }\nu }=0,\text{on}\mathrm{\Sigma },\end{array}\end{array}$(6.30) (6.31) $\begin{array}{rl}& \begin{array}{rl}& p_1+{\nu }_1\frac{\mathrm{\partial }{y}_1}{\mathrm{\partial }\nu }\ge 0,\frac{\mathrm{\partial }{y}_1}{\mathrm{\partial }\nu }[p_1+{\nu }_1\frac{\mathrm{\partial }{y}_1}{\mathrm{\partial }\nu }]=0,\text{on}\mathrm{\Sigma },\\ & p_2+{\nu }_2\frac{\mathrm{\partial }{y}_2}{\mathrm{\partial }\nu }\ge 0,\frac{\mathrm{\partial }{y}_2}{\mathrm{\partial }\nu }[p_2+{\nu }_2\frac{\mathrm{\partial }{y}_2}{\mathrm{\partial }\nu }]=0,\text{on}\mathrm{\Sigma },\end{array}\end{array}$(6.31) hence (6.32) $\begin{array}{rl}& u_1=\frac{\mathrm{\partial }{y}_1}{\mathrm{\partial }\nu }{|}_{\mathrm{\Sigma }},\\ & u_2=\frac{\mathrm{\partial }{y}_2}{\mathrm{\partial }\nu }{|}_{\mathrm{\Sigma }}.\end{array}$(6.32)

Example 6.4

Riemann–Liouville sense

We consider an example analogous to that considered in Example (6.1), but the fractional time derivative is considered in a Riemann–Liouville sense. The state equations are given by (6.33) $\begin{array}{rl}& {}_0^{ABR}{D}_t^{\alpha }{y}_1\left(u\right)-\mathrm{\Delta }{y}_1\left(u\right)+y_1\left(u\right)-y_2\left(u\right)\\ & =f_1\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.33) (6.34) $\begin{array}{rl}& {}_0^{ABR}{D}_t^{\alpha }{y}_2\left(u\right)-\mathrm{\Delta }{y}_2\left(u\right)+y_2\left(u\right)+y_1\left(u\right)\\ & =f_2\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_2\in L^2\left(Q\right),\end{array}$(6.34) (6.35) $I^{1-\alpha }{y}_1(x,0^{+};u)=y_{0,1}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(6.35) (6.36) $I^{1-\alpha }{y}_2(x,0^{+};u)=y_{0,2}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(6.36) (6.37) $\frac{\mathrm{\partial }{y}_1(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_1,\frac{\mathrm{\partial }{y}_2(x,t)}{\mathrm{\partial }{\nu }_A}{|}_{\mathrm{\Sigma }}=u_2,x\in \mathrm{\Gamma },t\in (0,T),$(6.37) where the fractional integral $I^{1-\alpha }$ and the derivative ${}_0^{ABR}{D}_t^{\alpha }$ are understand here in the Riemann–Liouville sense, $I^{1-\alpha }y(x,0^{+};u)=\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }y(x,t;u)$.

The adjoint state is given by (6.38) $-{}_t^{ABC}{D}_T^{\alpha }{p}_1\left(u\right)+\mathrm{\Delta }{p}_1\left(u\right)+p_1\left(u\right)+p_2\left(u\right)=0,\text{in}Q,$(6.38) (6.39) $-{}_t^{ABC}{D}_T^{\alpha }{p}_2\left(u\right)+\mathrm{\Delta }{p}_2\left(u\right)+p_2\left(u\right)-p_1\left(u\right)=0,\text{in}Q,$(6.39) (6.40) $\frac{\mathrm{\partial }{p}_1\left(u\right)}{\mathrm{\partial }\nu }=0,\frac{\mathrm{\partial }{p}_2\left(u\right)}{\mathrm{\partial }\nu }=0\text{on}\mathrm{\Sigma },$(6.40) (6.41) $p_1(x,T;u)=y_1\left(u\right)-z_d,\text{in}\mathrm{\Omega },p_2(x,T;u)=0\text{in}\mathrm{\Omega }.$(6.41) The optimality condition is given by (6.42) $\begin{array}{rl}& {\int }_{\mathrm{\Sigma }}\left[p_1\right(u\left)\right(v_1-u_1)+p_2(u\left)\right(v_2-u_2\left)\right]\mathrm{d}\mathrm{\Sigma }\\ & +(Nu,v-u{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^2}\ge 0,\mathrm{\forall }v\in U_{ad},u\in U_{ad}.\end{array}$(6.42)

Example 6.5

$n\times n$ system

We can generalize our results to n-dimensional coupled fractional system as follows. The state of the system is given, for each $i=1,2,\dots ,n$, by (6.43) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_i\left(u\right)-\mathrm{\Delta }{y}_i\left(u\right)+\sum _{j=1}^n{b}_{ij}{y}_j\left(u\right)\\ & =f_i\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.43) (6.44) $y_i(x,0;u)=y_{0,i}\left(x\right)\in L^2\left(\mathrm{\Omega }\right),x\in \mathrm{\Omega },$(6.44) (6.45) $\frac{\mathrm{\partial }{y}_i(x,t)}{\mathrm{\partial }\nu }{|}_{\mathrm{\Sigma }}=u_i,x\in \mathrm{\Gamma },t\in (0,T),$(6.45) (6.46) $b_{ij}=\left\{\begin{array}{ll}1,& i\ge j;\\ -1,& i<j.\end{array}\right.$(6.46)

The control $u=\{u_1,u_2,\dots ,u_n\}$ is taken in $\left(L^2\right(\mathrm{\Sigma }){)}^n$: (6.47) $u=\{u_1,u_2,\dots ,u_n\}\in U=\left(L^2\right(\mathrm{\Sigma }){)}^n.$(6.47) Problem (6.41(6.43) $\begin{array}{rl}& {}_0^{ABC}{D}_t^{\alpha }{y}_i\left(u\right)-\mathrm{\Delta }{y}_i\left(u\right)+\sum _{j=1}^n{b}_{ij}{y}_j\left(u\right)\\ & =f_i\left(t\right),\text{in}Q,a.e.t\in ]0,T[,f_1\in L^2\left(Q\right),\end{array}$(6.43) )–(6.44(6.46) $b_{ij}=\left\{\begin{array}{ll}1,& i\ge j;\\ -1,& i<j.\end{array}\right.$(6.46) ) admits a unique solution. To see this, we use the method developed in [15]: (6.48) $V=\left(H^1\right(\mathrm{\Omega }){)}^n,\phi =\{{\phi }_1,{\phi }_2,\dots ,{\phi }_n\}\in V,$(6.48) (6.49) $\begin{array}{rl}\pi (t;y,\phi )& =\pi (y,\phi )={\int }_{\mathrm{\Omega }}\sum _{i=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_i}{y}_i\left(x\right)\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_i}{\phi }_i\left(x\right)\mathrm{d}x\\ & +\sum _{i=1}^n{\int }_{\mathrm{\Omega }}\sum _{j=1}^n{b}_{i,j}{y}_j\left(x\right){\phi }_i\left(x\right)\mathrm{d}x.,\\ L\left(\phi \right)& =(f,\phi )={\int }_{\mathrm{\Omega }}\sum _{i=1}^n{f}_i(x,t){\phi }_i\left(x\right)\mathrm{d}x\\ & +{\int }_{\mathrm{\Gamma }}\sum _{i=1}^n{u}_i\left(t\right){\phi }_i\left(x\right)\mathrm{d}\mathrm{\Gamma }.\end{array}$(6.49) Let us consider the case where we have partial observation of the final state (6.50) $z\left(v\right)=y_1(x,T;v),$(6.50) and the cost function $J\left(v\right)$ for $v=\{v_1,v_2\}$ is given by (6.51) $\begin{array}{rl}J\left(v\right)& ={\int }_{\mathrm{\Omega }}\left(y_1\right(x,T;v)-z_d{)}^2\mathrm{d}x\\ & +(Nv,v{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^n},z_d\in L^2(\mathrm{\Omega }),\end{array}$(6.51) where $N=\{N_1,N_2,\dots ,N_n\}\in \mathcal{L}\left(\right(L^2\left(\mathrm{\Sigma }\right){)}^n,\left(L^2\right(\mathrm{\Sigma }\left){)}^n\right)$ is Hermitian positive definite operator: (6.52) $(Nu,u)\ge c||u|{|}_{L^2\left(\mathrm{\Sigma }\right)}^n,c>0.$(6.52) Control constraints: We define $U_{ad}$ (set of admissible controls) is closed, convex subset of $U=\left(L^2\right(\mathrm{\Sigma }){)}^n$.

Control problem: We want to minimize J over $U_{ad}$, i.e. find $u=\{u_1,u_2,\dots ,u_n\}$ such that (6.53) $J\left(u\right)=\underset{v=\{v_1,v_2,\dots ,v_n\}\in U_{ad}}{inf}J\left(v\right).$(6.53) The adjoint state is given by (6.54) $-{}_t^{ABC}{D}_T^{\alpha }{p}_i\left(u\right)+\mathrm{\Delta }{p}_i\left(u\right)+\sum _{j=1}^n{b}_{ji}{p}_j\left(u\right)=0,\text{in}Q,$(6.54) (6.55) $\frac{\mathrm{\partial }{p}_i\left(u\right)}{\mathrm{\partial }\nu }=0,\text{on}\mathrm{\Sigma },$(6.55) (6.56) $p_1(x,T;u)=y_1\left(u\right)-z_d,\text{in}\mathrm{\Omega },$(6.56) (6.57) $p_k(x,T;u)=0,k=2,3,\dots ,n,\text{in}\mathrm{\Omega },$(6.57) where $b_{ji}$ are the transpose of $b_{ij}$. The optimality condition is (6.58) $\begin{array}{rl}& {\int }_{\mathrm{\Sigma }}\sum _{i=1}^n{p}_i\left(u\right)(v_i-u_i)\mathrm{d}\mathrm{\Sigma }+(Nu,v-u{)}_{\left(L^2\right(\mathrm{\Sigma }){)}^n}\\ & \ge 0,\mathrm{\forall }v\in U_{ad},u\in U_{ad}.\end{array}$(6.58)

Remark 6.1

If we take $\alpha =1$ in the previews sections, we obtain the classical results in the optimal control with integer derivatives.

7. Conclusion

In this paper, we considered optimal control problem for coupled diffusion systems with Atangana–Baleanu derivatives and final observation. The analytical results were given in terms of Euler–Lagrange equations for the fractional optimal control problems. The formulation presented and the resulting equations are very similar to those for classical optimal control problems for coupled parabolic systems. The optimization problem presented in this paper constitutes a generalization of the optimal control problem of diffusion equations with Dirichlet boundary conditions considered in [12, 13, 15, 22–24, 35, 39] to coupled systems with Atangana–Baleanu time derivatives.

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

G. M. Bahaa http://orcid.org/0000-0001-6670-3110

A. Hamiaz http://orcid.org/0000-0001-5605-8728

Additional information

Funding

This work was supported by Taibah University.